Thomas Calculus 13th [Solutions]

420 Chapter 5 Integration
110.
8 8 8
24
8
1 1 1 y
1 y
1 1
dy 24 dy 24
4 2 2 4 2 2 2
16 4 4
sec 4
6sec 4
6 sec 2 sec 2
y y y y
4 2 4 2
6 6
3 4 12 2
111.
2/3
1
2/3 2
dy 3 dy 1 du , where u = 3y, du = 3 dy;
2/3 2 2/3 2 2 2
y 9 y 1 3 y (3 y) 1 u u 1
1 2 1 1
[sec u] sec 2 sec 2
2
3 4 12
2
y u 2, y 2 u 2
3 3
112.
6 / 5
1
6/ 5 5
6
dy dy 1 du,
2/ 5 2 2/ 5 2 2 2 2
2
y 5y 3 5y 5y 3 u u 3
where u 5 y, du 5 dy;
2
6
y u 2, y u 6
5 5
6
1 1 1 1 1 2 1 1 3 2
3
sec u
sec 2 sec
3 3 3 3 3 4 6 3 12 12 12 3 36
2
113. (a) 1
1
av( f )
1
1 ( 1) 1 ( mx b)
dx 2 1
2 2
mx
1 m(1) m( 1)
bx
b(1) b( 1) 1
2 2
1
2 2 2
2 (2 b)
(b) av( f 1
k
)
( ) 1
k ( k)
k mx b dx 2 k
2 2
mx
1 m( k) m( k)
bx
b( k) b( k)
2k
2
k
2k
2 2
1
2 (2 bk
k
) b
b
114. (a) yav
(b) yav
1
3 0
1
a 0
3
0
a
0
3x dx
ax dx
1
3
1
a
3 1/2 3 3/2
3
2
0 3 x dx 3 3
x 0
a 1/2 a 2 3/2
a
a x dx x
0 a 3 0
3 2 3/2 2 3/2 3
(3) (0) (2 3) 2
3 3 3 3
a 2 3/2 2 3/2 a
( a) (0)
2 a a 2 a
a 3 3 a 3 3
115. f av
1
b
( ) 1 b
[ ( )]
b a a
b a
( ) ( )
a
f b f a
[ f ( b) f ( a )]
so the average value of f over [ a, b ] is
b a b a
the slope of the secant line joining the points ( a, f ( a )) and ( b, f ( b )), which is the average rate of change
of f over [ a, b].
116. Yes, because the average value of f on [ a, b ] is 1
b
f ( x) dx . If the length of the interval is 2, then
b a a a 2
and the average value of the function is 1 b
( ) .
2 a f x dx
117. (a) d ( x ln x x C) x 1
dx
x
ln x 1 0 ln x
(b) average value 1
e
1 e
ln x dx [ x ln x x] 1 1 1
1 1
1 1 [( eln e e) (1ln1 1)] ( e e 1)
e e e 1 e 1 e 1
118.
average value 1
2
1
2
2 1 1 x dx ln x 1
ln 2 ln1 ln 2
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